Can someone provide me with some hint how to evaluate this limit? $$ \lim_{x \rightarrow 0}\left (\frac 1x- \frac 1{\sin x} \right ) $$ I tried l'hopital's rule but it didn't work.
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4Hint, combine the fractions (common denominator), then you can L'H. – Amzoti Jun 17 '13 at 00:16
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http://math.stackexchange.com/questions/400541/find-the-limit-without-use-of-lhopital-or-taylor-series-lim-limits-x-right – mnsh Jun 17 '13 at 08:16
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See What is the result of $\lim\limits_{x \to 0}(1/x - 1/\sin x)$? – Martin Sleziak Jun 17 '13 at 10:38
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The limit would be $\infty-\infty$ when $x\to0$.
As Amzoti have said in the comment, we can combine the fractions and get:
\begin{align} \lim_{x\to0}\frac{1}{x}-\frac{1}{\sin x}&=\lim_{x\to0}\frac{\sin x-x}{x\sin x},\text{now we can apply L'Hospital's Rule since the limit would be }\frac{0}{0} \\ \\ &=\lim_{x\to0}\frac{\cos x-1}{\sin x+x\cos x}\\ \\ &=\lim_{x\to0}\frac{-\sin x}{\cos x+\cos x-x\sin x}\\ \\ &=\boxed{0} \end{align}
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3If you put
$\sin x$instead of$sin x$, you'll get the more "operator-looking" $\sin x$. Also, you should have $-x\sin x$ in your last denominator, rather than $-x\cos x$. – Cameron Buie Jun 17 '13 at 00:23 -
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Rewrite the difference as (after Maclaurin series expansion and some algebra) $$ \frac{\sin x-x}{x \sin x}=\frac{x+O(x^3)-x}{x(x+O(x^3))} =\frac{O(x)}{1+O(x^2)}$$ which tends to $0$.
Alex
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